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That's basically the problem. I keep getting $\theta=90-\phi/2$. But I have a feeling its not right. What I did was draw line segments BD and AC. From there you get four triangles. I labeled the intersection of BD and AC as point P. From exterior angles I got my answer.

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3 Answers 3

Your answer is correct. The angle subtended at the centre is $2 \theta$, and by using angle sum property and the fact that tangents are perpendicular to the radii at the point of contact, we can obtain $2\theta + \phi = 180 \Rightarrow \theta = 90 - {\phi \over 2}$

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One way would be to let $E$ be the center of the circle. A standard result in geometry tells you that $AEC=2\theta$. And the two sides $AE$ and $CE$ are of equal lengths, and there are right angles at $A$ and $C$, and the sides $AD$ and $CD$ are also of equal lengths. So the triangle $EAD$ is right triangle congruent to $ECD$. One of the angles in that right triangle is $\theta$, so the other is $90^\circ-\theta$. Therefore what you're looking for is $180^\circ-2\theta$.

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You are correct. What makes you second-guess yourself? I think your method is sound. However, note that you are asked to express $\phi$ as a function of $\theta$, so you want

$$\theta = 90 -\dfrac{\phi}{2} \quad \iff \quad \phi = 180 - 2\theta$$

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Nice to see you, again angel here. Unfortunately, I have not been these days as an active one and honestly this makes me annoying. I hope I do the best for next days Amy. :-) –  Babak S. Mar 22 '13 at 20:07

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